What is the answer to solvefor y,x=(y^2+3)/(3y^2) ?

The answer to solvefor y,x=(y^2+3)/(3y^2) is y=sqrt(3/(3x-1)),y=-sqrt(3/(3x-1));\quad x\ne 1/3

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solvefor y,x=(y^2+3)/(3y^2)

solve for

y, x = \frac{y ^{2} + 3}{3 y ^{2}}

y = \frac{3}{3 x - 1}, y = - \frac{3}{3 x - 1}; x \neq = \frac{1}{3}

Solution steps

x = \frac{y ^{2} + 3}{3 y ^{2}}

Multiply both sides by

y^{2}

x y^{2} = \frac{y ^{2} + 3}{3}

Solve

x y^{2} = \frac{y ^{2} + 3}{3} : y = \frac{3}{3 x - 1}, y = - \frac{3}{3 x - 1}; x \neq = \frac{1}{3}

y = \frac{3}{3 x - 1}, y = - \frac{3}{3 x - 1}; x \neq = \frac{1}{3}

\frac{3}{6} = \frac{11}{a}

\frac{x + 2}{x + 3} + 1 = \frac{- 7}{x - 5}

\frac{4}{x + 2} = \frac{5}{2 x + 1}

\frac{4}{x + 7} = \frac{6}{x}

2 x - \frac{1}{x} = 1

What is the answer to solvefor y,x=(y^2+3)/(3y^2) ?
The answer to solvefor y,x=(y^2+3)/(3y^2) is y=sqrt(3/(3x-1)),y=-sqrt(3/(3x-1));\quad x\ne 1/3